int xe^-x dx

Question

int xe^-x dx

anonymous · Answer

$$\int\limits_{}^{}xe ^{-x}dx$$

callisto · Answer

$$\int xe^{-x}dx$$$$=-\int xd(e^{-x})$$$$=-(xe^{-x}-\int e^{-x}dx)$$$$=-[xe^{-x}-(- e^{-x})]+C$$$$= -xe^{-x}-e^{-x} +C$$$$= -e^{-x}( x+1)+C$$

anonymous · Answer

how do you get -$$\int\limits_{}^{}xd(e ^{-x}?$$

callisto · Answer

$$-\int xd(e^{-x}) =- \int x(-e^{-x})dx=\int xe^{-x}dx$$

anonymous · Answer

the int of e^-x equals -e^-x?

callisto · Answer

Yes.

anonymous · Answer

$$-e^{-x}(x-1)+C$$

should be the answer I think.

callisto · Answer

May I know where I made mistakes?

callisto · Answer

$$\int xe^{-x}dx$$$$=-\int xd(e^{-x})$$$$=-[xe^{-x}- \int e^{-x}dx]$$$$=-[xe^{-x}- (-e^{-x}) +C]$$$$=-(xe^{-x}+e^{-x}) +C$$$$=-e^{-x}(x+1) +C$$

callisto · Answer

I still get the same answer :|

zarkon · Answer

Callisto is correct

anonymous · Answer

in the third step callisto wrote $$-[xe ^{-x}-(-e ^{-x})+C]$$

where was the -(-e$$^{-x}$$ combined?

anonymous · Answer

mistyped that, but what I meant was the two negatives.

anonymous · Answer

what I meant was, when it's factored out, the -e^-x multiplied by the positive 1, gives negative e

anonymous · Answer

I'll look at it again. I'm sure I'm wrong.

anonymous · Answer

The signs don't match the previous step when I multiply the final answer.

zarkon · Answer

check by differentiating...
$$\frac{d}{dx}[-e^{-x}(x+1)+c]$$

$$=-e^{-x}+-e^{-x}(-1)(x+1)$$

$$=-e^{-x}+e^{-x}x+e^{-x}=xe^{-x}$$